# Is the function $f(x,y)=xy/(x^{2}+y^{2})$ where f(0,0) is defined to be 0 continuous?

Is the function $f(x,y)=xy/ (x^{2}+y^{2})$ where $f(0,0)$ is defined to be $0$ continuous? I don't think it is and I am trying to either show this by the definition or by showing that maybe a close set in $\mathbb{R}$ has an inverse set that is not closed in $\mathbb{R} ^{2}$. I tried the point $0$ but this is open in $\mathbb{R}$ Any hints or ideas? Thanks !

-

It is not continuous at $(0,0)$. Because $f(t,t)=1/2$ but $f(t,0)=0$, so if we approach to the origin along the line $y=x$ then $f(x,y)\to 1/2$ but if we approach to the origin along the x-axis then $f(x,y)\to 0$.

-
Nice, thank you. –  Jmaff Oct 8 '13 at 2:01

If you use polar coordinates

$$f(r\cos(\theta), r\sin(\theta))= {\sin(\theta)\cos(\theta)}=\frac{\sin(2\theta)}{2},$$

then, you can see that the limit has infinite number of values depending on the choice of theta which implies the limit does not exist. A related problem.

-

Consider an approach along the line $y = x$; then

$$f(x, y) = \frac{x^2}{x^2 + x^2} = \frac{1}{2}$$

for all $x \ne 0$. On the other hand, if we approach $(0, 0)$ along the line $y = 2x$,

$$f(x, y) = \frac{2x^2}{x^2 + 4x^2} = \frac{2}{5}$$

So there are two different paths toward the origin, each giving a different limit. Hence, the limit does not exist.

Or more simply, approach along $y = 0$.

-

Well, If it were continuous then for ALL $\epsilon >0$, there exist $\delta>0$ such that if $||x-a||<\delta$ then $||f(x)-f(a)||<\epsilon$. So it is enough to show one counter example. For a specific epsilon, say $\epsilon =\frac{1}{4}$, then clearly $$||(\frac{\delta}{2}, \frac{\delta}{2})-(0,0)|| = ||(\frac{\delta}{2},\frac{\delta}{2})|| = \frac{\delta}{\sqrt{2}}<\delta$$

But $$||\frac{\frac{\delta}{2}\frac{\delta}{2}}{\frac{\delta^2}{4}{}\frac{\delta^2}{4}}- 0||=\frac{1}{2}>\frac{1}{4} = \epsilon$$ There you go you have a counter example here where we picked a specific epsilon and could not conclude $||f(x)-f(a)||<\epsilon$

The above answers are correct but you should be able to prove this using delta epsilon definition like I did. Note that the answers above say you approach the origin along y=x suggests the delta I pick for the proof. You see I picked $(\frac{\delta}{2}, \frac{\delta}{2})$ in the form of $(x,x)$ for my $(x,y)$ since I'm approaching the origin along y=x.

-